Question

A stone of mass 1 kg is thrown with a velocity of 20 m/s across the frozen surface of lake and comes to rest after travelling a distance of 50 M .what is the force of friction between the stone and the ice.​

Answer 1

f=ma

u=20m/s

v= 0m/s

s=d/t

20=50/t

t=50/20=2.5sec

a=v-u/t

a=0-20/2.5

a= -20/2.5

a=-8 m/s^2

f= 1* - 8

f= -8 N

so f is - 8N

HOPE THIS HELPED

Answer 2

☺ Hello mate__ ❤

◾◾here is your answer...

u = 20 m/s

v = 0 m/s

s = 50 m

According to the third equation of motion:

v^2 = u^2 + 2as

(0)^2 = (20)^2 + 2 × a × 50

a = – 4 m/s2

★The negative sign indicates that acceleration is acting against the motion of the stone.

m = 1 kg

From Newton's second law of motion:

F = Mass x Acceleration

F= ma

F= 1 × (– 4) = – 4 N

Hence, the force of friction between the stone and the ice is – 4 N.

I hope, this will help you.

Thank you______❤

✿┅═══❁✿ Be Brainly✿❁═══┅✿